P2885 [USACO07NOV]电话线Telephone Wire

2021年09月15日 阅读数:3
这篇文章主要向大家介绍P2885 [USACO07NOV]电话线Telephone Wire,主要内容包括基础应用、实用技巧、原理机制等方面,希望对大家有所帮助。

题目描述

Farmer John's cows are getting restless about their poor telephone service; they want FJ to replace the old telephone wire with new, more efficient wire. The new wiring will utilize N (2 ≤ N ≤ 100,000) already-installed telephone poles, each with some heighti meters (1 ≤ heighti ≤ 100). The new wire will connect the tops of each pair of adjacent poles and will incur a penalty cost C × the two poles' height difference for each section of wire where the poles are of different heights (1 ≤ C ≤ 100). The poles, of course, are in a certain sequence and can not be moved.html

Farmer John figures that if he makes some poles taller he can reduce his penalties, though with some other additional cost. He can add an integer X number of meters to a pole at a cost of X2.ios

Help Farmer John determine the cheapest combination of growing pole heights and connecting wire so that the cows can get their new and improved service.less

给出若干棵树的高度,你能够进行一种操做:把某棵树增高h,花费为h*h。ide

操做完成后连线,两棵树间花费为高度差*定值c。post

求两种花费加和最小值。spa

输入输出格式

输入格式:rest

 

  • Line 1: Two space-separated integers: N and Ccode

  • Lines 2..N+1: Line i+1 contains a single integer: heighti

 

输出格式:htm

 

  • Line 1: The minimum total amount of money that it will cost Farmer John to attach the new telephone wire.

 

输入输出样例

输入样例#1:
5 2
2
3
5
1
4
输出样例#1:15




一开始本身写了个DP,竟然能过样例
特别兴奋,
可是交上去只有70分
发现时间复杂度有点高
思路比较简单:
咱们能够很容易的看出这道题具备无后效性,
用dp[i][j]表示前i棵树,第i棵树高度为j的最小代价
先预处理一下dp[1][j],而后对于每一棵树,咱们枚举它的前一棵树的高度和这棵树的高度,
计算一下就好
时间复杂度n*h*h

P2885 [USACO07NOV]电话线Telephone Wire_c语言 P2885 [USACO07NOV]电话线Telephone Wire_题型_02
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<cstdlib>
 9 using namespace std;
10 const int MAXN=100001;
11 const int INF =0x7f7f7f7f;
12 inline void read(int &n)
13 {
14     char c='+';bool flag=0;n=0;
15     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
16     while(c>='0'&&c<='9')n=n*10+c-48,c=getchar();
17 }
18 int dp[MAXN][201];// 第i棵树,高度为j的最小花费
19 int n,C; 
20 int a[MAXN];
21 int maxheight;
22 int main()
23 {
24     read(n);read(C);
25     memset(dp,INF,sizeof(dp));
26     for(int i=1;i<=n;i++)
27         read(a[i]),maxheight=max(maxheight,a[i]);
28     for(int i=a[1];i<=maxheight;i++)
29         dp[1][i]=(i-a[1])*(i-a[1]);
30         
31     for(int i=2;i<=n;i++)//枚举全部树 
32         for(int j=a[i];j<=maxheight;j++)//枚举这棵树的高度 
33             for(int k=a[i-1];k<=maxheight;k++)//枚举前一棵树的高度 
34                 dp[i][j]=min(dp[i][j],
35                             ((j-a[i])*(j-a[i])+(dp[i-1][k])+abs(j-k)*C));
36                             
37                             
38     int ans=0x7fffff;
39     for(int i=a[n];i<=maxheight;i++)
40         ans=min(ans,dp[n][i]);
41     printf("%d",ans);
42     return 0;
43 }
View Code

而后化简一下式子

 

 
    
P2885 [USACO07NOV]电话线Telephone Wire_c语言 P2885 [USACO07NOV]电话线Telephone Wire_题型_02
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 using namespace std;
 6 const int MAXN=300005;
 7 const int INF =0x7fffff;
 8 const int maxheight=100;
 9 int dp[301];// 第i棵树,高度为j的最小花费
10 int f[301];
11 int n,C;
12 int a[MAXN];
13 int bgsum[MAXN];
14 int edsum[MAXN];
15 int main() {
16     scanf("%d%d",&n,&C);
17     for(int i=0; i<n; i++)
18         scanf("%d",&a[i]);
19     memset(dp, 0x3f, sizeof(dp));
20     for(int i=a[0]; i<=maxheight; i++)
21         dp[i]=(i-a[0])*(i-a[0]);
22 
23     for(int i=1; i<n; i++) { //枚举全部树
24         memcpy(f,dp,sizeof(dp));
25         for(int j=0; j<=maxheight; j++)    dp[j]=bgsum[j]=edsum[j]=INF;
26 
27         bgsum[0]=f[0];
28         for(int j=1; j<=maxheight; j++)
29             bgsum[j]=min(bgsum[j-1],f[j]-C*j);
30 
31         edsum[maxheight]=f[maxheight]+maxheight*C;
32         for(int j=maxheight-1; j>=0; j--)
33             edsum[j]=min(edsum[j+1],f[j]+C*j);
34 
35         for(int j=a[i]; j<=maxheight; j++) //枚举这棵树的高度
36             dp[j]=min(edsum[j]-C*j,bgsum[j]+C*j)+(j-a[i])*(j-a[i]);
37     }
38     int ans=0x7fffff;
39     for(int i=a[n-1]; i<=maxheight; i++)
40         ans=min(ans,dp[i]);
41     printf("%d",ans);
42     return 0;
43 }
View Code